Implementing the '<' operator

Question

What is the correct way to implement the '<' operator for using the following simple class as a key in a std::map (functions are omitted)?

class Person
{
public:
   bool valid = false;
   std::string name = "";
   std::string id = "";
}

not a dupe but a good reference: http://stackoverflow.com/questions/4421706/operator-overloading — NathanOliver, Nov 14 '16 at 21:02
You havea field named `id`. Its name implies that it is unique. Is it? If so, you could make use of that fact. — , Nov 14 '16 at 21:04
There is no "the correct" way. It is up to you really, as long as it satisfies *strict weak ordering*. — juanchopanza, Nov 14 '16 at 22:46

score 4 · Accepted Answer · edited May 23 '17 at 12:07

4

You could use std::tie.

class Person
{
public:
    bool valid = false;
    std::string name = "";
    std::string id = "";

    bool operator<(const Person& r) const
    {
        return std::tie(valid, name, id) < std::tie(r.valid, r.name, r.id);
    }
}

Explanation:

std::tie(xs...) creates an std::tuple of references to the passed xs... arguments. Comparison between two std::tuple instances works by lexicographically comparing the elements, thus providing an ordering for your type.

More information here on cppsamples and in this question.

edited May 23 '17 at 12:07

Community

1
1

answered Nov 14 '16 at 21:01

Vittorio Romeo

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2

@RichardHodges: now let's fight for **member function** vs **free function**! – Vittorio Romeo Nov 14 '16 at 21:04
Better yet explain how/why `tie` is the way to go/what it is doing. – NathanOliver Nov 14 '16 at 21:05
always free functions for binary ops :) – Richard Hodges Nov 14 '16 at 21:08
@NathanOliver: added a brief explanation. – Vittorio Romeo Nov 14 '16 at 21:09

R Sahu · Answer 2 · 2016-11-14T21:15:04.700

4

You can use std::tie as suggested by the other answers. If you want to clearly see the logic in your own function or don't have access to a C++11 compiler, you can implement it as:

class Person
{
   public:
      bool valid = false;
      std::string name = "";
      std::string id = "";

      bool operator<(Person const& rhs) const
      {
         if ( this->valid != rhs.valid )
         {
            return ( this->valid < rhs.valid );
         }
         if ( this->name != rhs.name )
         {
            return ( this->name < rhs.name );
         }
         return ( this->id < rhs.id );
      }
};

edited Nov 14 '16 at 21:15

answered Nov 14 '16 at 21:10

R Sahu

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If it were me, I'd make it so all invalid objects are equivalent without relying on the other members. That's something you can't do with `std::tie`. – Mark Ransom Nov 14 '16 at 21:12
1

It should be noted, that `std::tuple` uses equivalency, while this implementation uses equality. It's the same for these types, but not in general. – StoryTeller - Unslander Monica Nov 14 '16 at 21:14

score 2 · Answer 3 · answered Nov 14 '16 at 21:02

#include <string>
#include <tuple>

class Person
{
public:
   bool valid = false;
   std::string name = "";
   std::string id = "";
};

bool operator<(const Person& l, const Person& r)
{
    return std::tie(l.valid, l.name, l.id) < std::tie(r.valid, r.name, r.id);
}

Implementing the '<' operator

3 Answers3